Question

# A ball is projected horizontally from top of a $$80$$ m deep well with velocity $$10$$ m/s. Then particle will fall on the bottom at a distance of (all the collisions with the wall are elastic)

A
5 m from A
B
5 m from B
C
2 m from A
D
2 m from B

Solution

## The correct options are B $$5$$ m from B C $$2$$ m from ATotal time taken by the ball to reach at bottom $$=\sqrt{\dfrac{2H}{g}}=\sqrt{\dfrac{2\times 80}{10}}=4\ sec$$Let time taken in one collision is $$t_{collision}$$. Then $$t_{collision} \times 10 = 7$$ since horizontal velocity is 10 m/s$$\Rightarrow t_{collision}=0.7$$ sec.No. of collisions $$=\dfrac{4}{.7}=5\dfrac{5}{7}$$ (5th collisions from wall B)Horizontal distance traveled in between $$2$$ successive collisions $$=7\:m$$$$\therefore$$Horizontal distance traveled in $$5/7$$ part of collisions $$=$$ The 1st collision will be with wall B, 2nd collision with wall A, 3rd collision with wall B, 4th collision with wall A and  5th collision will be with wall B.  After 5th collision, the ball   will cover a distance of $$\dfrac{5}{7}\times 7=5 m$$ from wall B i.e it will have collision at a distance of 2m from wall A.Physics

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